Problem: The polynomial $p(x)=x^3+7x^2-36$ has a known factor of $(x+3)$. Rewrite $p(x)$ as a product of linear factors. $p(x)=$
Solution: We know $(x+3)$ is a factor of $p(x)$. This means that $p(x)=(x+3)\cdot q(x)$ for some polynomial $q(x)$. We can find $q(x)$ using polynomial division, and then we can factor $q(x)$. This way, we will be able to rewrite $p(x)$ as a product of linear factors. Dividing $p(x)$ by $(x+3)$ Notice that $p(x)$ is missing a $1^{\text{st}}$ degree term. Let's add it as $0x$. $\begin{array}{r} x^2+\phantom{1}4x-12 \\ x+3|\overline{x^3+7x^2+\phantom{1}0x-36} \\ \mathllap{-(}\underline{x^3+3x^2\phantom{+10x-36}\rlap)} \\ 4x^2+\phantom{1}0x-36 \\ \mathllap{-(}\underline{4x^2+12x\phantom{-36}\rlap)} \\ -12x-36 \\ \mathllap{-(}\underline{-12x-36\rlap)} \\ 0 \end{array}$ We find that $q(x)=x^2+4x-12$. Factoring $q(x)$ We can factor $x^2+{4}x{-12}$ as $(x+m)(x+n)$ where $m+n={4}$ and $m\cdot n={-12}$. Such numbers are $6$ and $-2$, so the factored expression is $(x+6)(x-2)$. Putting it all together $\begin{aligned} p(x)&=x^3+7x^2-36 \\\\ &=(x+3)(x^2+4x-12) \\\\ &=(x+3)(x+6)(x-2) \end{aligned}$